First of all, I don't like the problem "statement" at all, it makes little mathematical sense. Did you get that out of a textbook, or did you make it up ?

The ("Fermat-like") equation [tex]a^n + b^n = c^n[/tex] has infinitely many real solution sets (a,b,c) for every positive integral value of n. Asking what solutions exist when n is infinite, or even "n tends to infinity" is meaningless. This is not evaluating the limit of an expression, this is gobbledegook.

Explain what mathematical principal I'm violating here? Because from what I see above I have every right asking the question I did.

As has been discussed a very large number (but still finite) number of times in this forum, infinity is NOT a number. You cannot set up an equation using infinity and expect it to have a solution.

If, OTOH, you're asking whether that equation has solutions for very large finite values of n, then the answer is always YES, so the premise you want to prove is wrong.

When you are evaluating the limits of expressions as something "tends" to infinity, all you are doing is investigating the behaviour or a mathematical expression as one of the terms is made arbitrarily large. Note that even if the limit exists, the expression can NEVER actually equal the limit. It can only approach it.

Just as you cannot speak of the expression ever having the ACTUAL value of the limit, you cannot set up an equation using those rules.

As has been discussed a very large number (but still finite) number of times in this forum, infinity is NOT a number. You cannot set up an equation using infinity and expect it to have a solution.

Oh? [tex]lim_{n\rightarrow\infty}\frac{1}{n}=0[/tex]

Curious3141 said:

If, OTOH, you're asking whether that equation has solutions for very large finite values of n, then the answer is always YES, so the premise you want to prove is wrong.

Of course there's a solution for every finite n. So?

Curious3141 said:

When you are evaluating the limits of expressions as something "tends" to infinity, all you are doing is investigating the behaviour or a mathematical expression as one of the terms is made arbitrarily large.

So I'm investigating the value of the expression as a variable is increasing without bounds. And?

Curious3141 said:

Just as you cannot speak of the expression ever having the ACTUAL value of the limit, you cannot set up an equation using those rules.

By your logic, suppose we have [tex]y=\frac{1}{x}[/tex]. We find that there is a non-zero [tex]y[/tex] for any arbitrarily large [tex]x[/tex], and therefore makes no sense to ask whether there's a limit for [tex]y=\frac{1}{x}[/tex] as [tex]x\rightarrow\infty[/tex].

To Illustrate: a case where you could use the limit is the case of the Fibonacci numbers: 1,1,2,3,5,8,13,21,34,55,89,144....where the series in continued by the formula: F(N-1)+F(N)=F(N+1), or we add two successive numbers to get the next one.

Now it happens to be that this series is closely connected with the Golden Mean:
[tex]\frac{-1+\sqrt(5)}{2}[/tex] =.618034...

We look at: 34/55 =.618182..;55/89 =.617978...;89/144 =.618056...

Thus we might wonder if the limit as n goes to infinity of F(n)/F(n+1) = Golden Mean...which it does!

This limit is so good, that if divided 144 by the Golden Mean we get 232.996, which rounds off to 233 =89+144. (In fact it is good even from the second "1" giving 1/GM = 1.6180 rounds to 2. 2/GM =3.23 rounds to 3, 3/GM=4.85 rounds to 5!

Oh-what ? [tex]\frac{1}{n}[/tex] is an EXPRESSION, not an EQUATION. And it *never* equals zero no matter what value of n you choose. Only the limit is zero. There is a fundamental distinction here you don't seem to be grasping.

Of course there's a solution for every finite n. So?

So the premise you set about to "prove" is WRONG !!

I'm investigating the value of the expression as a variable is increasing without bounds. And?

No, you are NOT. You are investigating an EQUATION as n tends to infinity. Which makes no sense.

You CAN investigate the behavior of the EXPRESSION [tex]{(a^n + b^n)}^{\frac{1}{n}}[/tex] as n tends to infinity. If, for simplicity, we allow only for positive real values of a and b, then